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3 years ago

An object will not change its motion unless an outside force acts on it true or false

Physics

2 answers:

Gre4nikov [31]3 years ago

7 0

Answer:

True!

Explanation:

Newton's First Law :)

Komok [63]3 years ago

6 0

Answer:

True

Explanation:

Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.

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"KATZPSEF1 7.P.053.MI. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Two black holes (the remains of exploded stars), separated by

eduard

Answer:

There are two possible solutions.

M1 = 4.68*10^30kg, M2 = 5.53*10^30kg

M1 = 5.53*10^30kg, M2 = 4.7*10^29kg

Explanation:

In order to find the mass of each black hole, you take into account the gravitational force between them and the sum of their masses.

You use the formula for the gravitational force between two masses:

$F_g=G\frac{M_1M_2}{r^2}$ (1)

G: Cavendish's constant = 6.674*10^-11 m^3/kg.s^2

M1, M2: mass of each black hole = ?

r: distance between the black holes = 10.0 AU = 10.0(1.50*10^11m) = 1.5*10^12m

Fg: gravitational force between the black holes = 7.70*10^25N

Furthermore, you take into account that the sum of the masses M1 and M2 is:

M1 + M2 = 6.00*10^30 kg (2)

You solve the equation (2) for M2.

$M_2=6.00*10^{30}-M_1$

Next, you replace the obtained expression for M2 into the equation (1) and solve for M1, as follow (for simplicity, you do not add the units):

$F_g=G\frac{M_1(6.00*10^{30}-M_1)}{r^2}\\\\\frac{r^2F_g}{G}=6.00*10^{30}M_1-M_1^2\\\\\frac{(1.5*10^{12})^2(7.70*10^{25})}{(6.674*10^{-11}}=6.00*10^{30}M_1-M_1^2\\\\2.59*10^{60}=6.00*10^{30}M_1-M_1^2\\\\M_1^2-6.00*10^{30}M_1+2.59*10^{60}=0$

Then, you have obtained a quadratic polynomial. You solve it with the quadratic formula:

$M_1=\frac{-(-6.00*10^{30})\pm \sqrt{(-6.00*10^{30})^2-4(1)(2.59*10^{60}))}}{2(1)}\\\\M_1=\frac{6.00*10^{30}\pm 5.06*10^{30}}{2}\\\\M_1=4.68*10^{29}\\\\M_1=5.53*10^{30}$

Both results are consistent, then the mass of one black hole can be 4.68*10^30kg and also 5.53*10^30kg.

The other black hole has a mass of:

$M_2=6.00*10^{30}kg-4.68*10^{29}kg=5.53*10^{30}kg\\\\M_2=6.00*10^{30}kg-5.53*10^{30}kg=4.7*10^{29}kg$

Hence, you have a pair of solutions:

M1 = 4.68*10^30kg, M2 = 5.53*10^30kg

M1 = 5.53*10^30kg, M2 = 4.7*10^29kg

3 0

3 years ago

An object stays at rest until what happens to it?

bixtya [17]

D. an outside or unbalanced force acts upon the object.

4 0

3 years ago

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

initial temperature of metals, = $T_m$
initial temperature of water, = $T_i$
specific heat capacity of copper, $C_p$ = 0.385 J/g.K
specific heat capacity of aluminum, $C_A$ = 0.9 J/g.K
both metals have equal mass = m

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

The change in heat energy for copper metal;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

For aluminum metal, the quantity of heat transferred is calculated as;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

The change in heat energy for aluminum metal ;

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

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6 0

3 years ago

Elaborate on the suitability of "cola" type drinks to polish chrome surfaces.

nexus9112 [7]

The answer is the phosphoric acid in cola easily removes dirt and grime. The "cola" type drinks is suitable to polish chrome surfaces. Cola contains a very diluted solution of phosphoric acid, which can remove rust, dirt, and grime from chrome surface.

3 0

4 years ago

A hummingbird flies 2.0 m along a straight path at a height of 5.3 m above the ground. Upon spotting a flower below, the humming

stepan [7]

To solve this problem we will apply the concepts of Pythagoras to find the net path. An easy way is to graph the path to trace the path, which resembles that of a right triangle. The information provided is equivalent to that of the two short sides of the triangle, so we will find the longest side by the Pythagorean theorem. In this way,

$OC = \sqrt{2^2+2.4^2}$

$OC = 3.12m$

Therefore the magnitude of the hummingbird’s total displacement is 3.12m

6 0

4 years ago